作业帮一道极限题（关联三角函数）lim（x—>π/3)(sinx-sinπ/3)/(x-π/3)lim(x—>0)(cosx-1)/x为啥我算这总是无穷大?
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一道极限题（关联三角函数）lim（x—>π/3)(sinx-sinπ/3)/(x-π/3)lim(x—>0)(cosx-1)/x为啥我算这总是无穷大?
一道极限题（关联三角函数）
lim（x—>π/3)(sinx-sinπ/3)/(x-π/3)
lim(x—>0)(cosx-1)/x
为啥我算这总是无穷大?
一道极限题（关联三角函数）lim（x—>π/3)(sinx-sinπ/3)/(x-π/3)lim(x—>0)(cosx-1)/x为啥我算这总是无穷大?
罗比塔法则
1.cosπ/3=1/2
2.0
也可直接做
lim（x—>π/3)(sinx-sinπ/3)/(x-π/3)
=lim（x—>π/3)2sin(x/2-π/6)cos(x/2+π/6)/(x-π/3)
=lim（x—>π/3)2sin(x/2-π/6)cos(x/2+π/6)/2(x/2-π/6)
因为x—>0,sinx/x=1
则
sin(x/2-π/6)/(x/2-π/6)=1
lim（x—>π/3)2sin(x/2-π/6)cos(x/2+π/6)/2(x/2-π/6)
=lim（x—>π/3)cos(x/2+π/6)
=cosπ/3
=1/2
lim(x—>0)(cosx-1)/x
=lim(x—>0)-2(sinx/2)^2/x
因为x—>0,sinx/x=1
lim(x—>0)-2(x/2)^2/x
=0
用洛必干塔法则:
lim（x—>π/3)(sinx-sinπ/3)/(x-π/3)=lim（x—>π/3)cosx=1/2
lim(x—>0)(cosx-1)/x=lim(x—>0)sinx=0
(cosx-1)/x
=-2[sin(x/2)]^2/x ->0, x->0
第一个式子可令y=x-π/3,然后展开得
(sinx-sinπ/3)/(x-π/3)
=[sinycos(π/3)+cosysin(π/3)-sin(π/3)]/y ->1/2, y->0
你算出无穷大，估计忘了分子上还要减去sin(π/3)
利用洛必达法则~~分子分母分别就导可知：
1. cosπ/3
2. 0
lim（x—>π/3)(sinx-sinπ/3)/(x-π/3)的实际含义是函数在π/3的切线的斜率，∴ 原式=cosπ/3=0.5
第二个式子表示lim(x—>0)(cosx-cos0)/(x-0).也就是函数y=cosx在x=0处的导数值，∴原式=cos0=1